Church_statics and Dynamics 1886

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3/216

Cornell

University Library

TA

350.C563

1886

Statics

and dynamics for

engineering

stu

3

1924

004

689 166


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9/216

STATICS

AND

DYNAMICS

FOR

ENGINEERING

STUDENTS.

IRVING

P.

CHURCH,

C.E.,

ASST.

PROFESSOR

OF

CIVIL ENGINEERING,

CORNELL

UNIVERSITY.

SECOND

EDITION.

NEW YORK:

JOHN

WILEY

&

SONS.

1886.


10/216

UNIVERSITY

\LIBRARY

Copyright,

1886,

By

IRVING

P.

CHURCH.

.

Press of

J. J.

Little &

Co.,

Nos. 10 to

20

Astor

Place,

New

York.


11/216

PREFACE.

For

the

Engineering

student,

pursuing the

study

of

Applied

Mechanics

as

part of

his

professional

training

and not

as ad-

ditional

mathematical

culture,

not

only

is

a

thoroughly

system-

atic, clear, and logical

treatment

of the subject

quite essential,

but

one

which

presents

the

quantities

and

conceptions

involved

in

as

practical

and concrete

a

form as

possible,

with

all the

aids

of the printer's

and

engraver's

arts

;

and

especially

one

which,

besides showing

the

derivation

of formulae

from

principles,

in-

culcates,

illustrates,

and lays

stress

on

correct

numerical

substi-

tution,

and

the

proper

use

of

units

;

for

without

this

no

reliable

numerical

results can be

reached,

and the

principal

object of

those

formulae

is frustrated.

With these

requirements

in

view,

an

experience

of ten

years

in

teaching

the Mechanics

of

Engineering at

this

institution

has

led the

writer to prepare

the present work,

embodying

the

fol-

lowing special

features :

1.

The diagrams are very

numerous

(about

one

to every

page ;

an

appeal

to

the

eye

is

often

worth

a

page

of description).

2.

The

diagrams are very

full and

explicit,

thus

saving

time

and

mental

effort

to

the

student.

In problems

in

Dynamics

three kinds

of

arrows are used, to

distinguish

forces, velocities,

and

accelerations, respectively.

3.

Illustrations and

examples of a

practical

nature,

both

algebraic

and

numerical, are of

frequent

occurrence.

4.

Formulae

are

divided

into two classes

;

those (homo-

geneous)

admitting the

use

of

any

system

of

units

whatever

for

measurements

of

force,

space, mass,

and

time,

in

numerical

substitution

;

and those

which are

true only

for specified units.

Attention is repeatedly

called

to the

matter of correct

numerical

substitution,

especially

in

Dynamics, where

time

and

mass,

as

well as

force

and

space,

are

among

the

quantities considered.

5.

The

general theorem

of Work

and Energy in

machines

is

developed

gradually

by

definite and

limited steps,

in

preference

to

giving

a

single

demonstration

which,

from

its

generality,

might

be

too vague

and

abstruse to

be

readily

grasped

by

the

student.

6.

As

to

the vexed

question

of

Centrifugal

force,

a con-

sistent

application of

Newton's

Laws

is made,

necessitating

the

result

that

the

centrifugal

force,

properly

so

called,

is

not

an

action

(force)

upon

the

body

constrained

to

move

in

the

curve,

but

upon

the

constraining

body.

7.

The

definition

of

force is

made to

include

and illustrate

Newton's

law

of

action

and

reaction,

the

misunderstanding of

which

leads

to

such

lengthy

and

unprofitable

discussions

in

technical

journals

every few

years.

Cobnell

University,

Ithaca,

N.

T.,

February,

1886.


12/216


13/216

TABLE

OF

CONTENTS.

PRELIMINARY

CHAPTER.

PAGE

1-15(1.

Definitions. Kinds

of Quantity.

Homogeneous Equa-

tions.

Parallelogram

of

Forces

1

PAET

I. STATICS.

Chapter I.

Statics

op a Material

Point.

16-19.

Composition

and Equilibrium

of

Concurrent

Forces

8

Chapter II.

Parallel Forces

and the Centre

op

Gravity.

20-22.

Parallel

Forces

13

23-27*. Centre

of Gravity.

Problems.

Centrobaric

Method.

. . 18

Chapter

III.

Statics op a Rigid Body.

28-34.

Couples

27

35^-39. Composition

and

Equilibrium

of Non-concurrent

Forces.

31

Chapter IV.

Statics

of

Flexible Cords.

40-48.

Postulates.

Suspended

Weights. Parabolic

Cord.

Cat-

enary

42

PAET

II.

DYNAMICS.

Chapter I.

Rectilinear

Motion

op a

Material Point.

49-55.

Uniform

Motion.

Falling Bodies.

Newton's

Laws.

Mass

49

56-60.

Uniformly

Accelerated

Motion.

Graphic

Representa-

tions.

Variably

Accelerated

Motions.

Impact...

54

Chapter

II.

Virtual

Velocities.

61-69.

Definitions

and

Propositions.

Connecting-rod.

Prob-

lems

67


14/216

VI

CONTENTS.

Chapter III.

Curvilinear

Motiox op

a

Material

Point.

PAGE

70-74.

Composition of Motions,

of Velocities,

etc.

General

Equations

72

75-84,

Normal Acceleration.

Centripetal

and

Centrifugal

Forces. Simple

Pendulums.

Projectiles.

Rela-

tive Motion

;

77

Chapter

IV.

Moment

of

Inertia.

85-94.

Plane

Figures.

Rigid

Bodies.

Reduction

Formulae.

The Rectangle,

Triangle, etc.

Compound

Plane

Figures. Polar Moment of

Inertia

91

95-104. Rod.

Thin

Plates.

Geometric

Solids

98

105-107. Numerical Substitution. Ellipsoid of

Inertia

102

Chapter

V.

Dynamics of

a Rigid

Body.

108-115.

Translation. Rotation about

a

Fixed Axis. Centre of

Percussion

105

116-121.

Torsion

Balance. Compound Pendulum. The Fly-

wheel 116

122-123.

Uniform

Rotation.

Centrifugal

Action.

Free

Axes. 125

124-126.

Rolling

Motions. Parallel

Rod

of

Locomotive

130

Chapter

VI.

Work, Energy, and Power.

127-134.

Work.

Power.

Horse-power.

Kinetic

Energy

133

135-138.

Steam-hammer.

Pile-driving.

Inelastic Impact

138

139-141.

Rotary

Motion. Equivalent

Systems

of

Forces. Any

Motion

of

a

Rigid

Body

143

142-146.

Work

and

Kinetic

Energy

in

a

Moving

Machine

of

Rigid

Parts

147

147-155. Power

of Motors.

Potential Energy. Heat,

etc.

Dy-

namometers. Atwood's Machine.

Boat-rowing.

Examples

153

Chapter

VII.

Friction.

156-164.

Sliding

Friction. Its Laws.

Bent

Lever

164

165-171.

Axle

-

friction.

Friction Wheels.

Pivots.

Belting.

Transmission

of

Power

by

Belting

,

175

172-177.

Rolling

Friction.

Brakes.

Engine-friction.

Anoma-

lies

in Friction. Rigidity

of

Cordage.

Examples.

186


15/216

MECHANICS

OF

ENGINEERING.

PRELIMINARY

CHAPTER.

1.

Mechanics

treats

of the

mutual

actions and

relative mo-

tions

of material

bodies,

solid,

liquid,

and

gaseous

; and

by

Mechanics

of

Engineering

is

meant

a presentment

of

those

principles of pure mechanics, and their

applications,

which

are

of

special service

in

engineering

problems.

2.

Kinds

of Quantity.

Mechanics

involves

the

following

fundamental

kinds

of

quantity

:

Space,

of

one, two,

or

three

dimensions,

i.e., length, surface,

or

volume, respectively

; time,

which

needs

no definition

here;

force

and

mass,

as defined

be-

low

;

and

abstract numbers,

whose values

are

independent

of

arbitrary

units,

as,

for

example,

a

ratio.

3.

Force.

A

force

is one

of a

pair

of

equal,

opposite,

and

simultaneous

actions between two bodies,

by

which

the

state

of

their

motions is

altered

or a

change

of

form in the

bodies

themselves

is

effected.

Pressure,

attraction,

repulsion,

and

traction

are

instances in

point.

Muscular sensation conveys

the

idea of

force,

while

a

spring-balance

gives an

absolute

measure

of

it,

a beam-balance

only a

relative

measure.

In

accordance

with

Newton's

third law of motion, that action

and

reaction are

equal,

opposite,

and

simultaneous,

forces

always

occur

in

pairs; thus,

if a

pressure

of 40

lbs. exists

between

bodies A

and

B,

if A

is considered by

itself

(i.e.,

free ),

apart

from

all

other bodies

whose

actions upon

it

are

called

forces,

among

these

forces

will

be

one of

40

lbs. directed from

B

toward A.

Similarly,

if

B

is

under

consideration, a

force


16/216

2

MECHANICS

OF

ENGINEERING.

of

4:0

lbs.

directed

from

A

toward

B

takes its place

among

the

forces

acting

on

B.

This

is

the

interpretation

of

Newton's

third law.

In

conceiving

of

a

force

as

applied at a

certain

point

of

a

body

it is useful to imagine

one end

of

an imponderable spiral

spring

in a

state

of compression

(or

tension)

as

attached at the

given

point,

the

axis

of

the spring

having

the

given

direction

of

the force.

4.

Mass

is the

quantity of matter in

a

body.

The

masses

of

several bodies

being

proportional

to their

weights

at

the same

locality on the

earth's surface,

in

physics

the

weight

is

taken

as the mass, but in practical

engineering another

mode

is

used

for measuring

it

(as

explained in

a

subsequent chapter),

viz.:

the mass

of a body

is equal to

its

weight divided by

the

ac-

celeration of

gravity in

tlie

locality

where

the

weight

is

taken,

or,

symbolically,

_3

=

G

-s-

g.

This

quotient is

a

constant

quantity,

as

it

should

be,

siuce

the

mass

of a body is

invariable

wherever the

body be

carried.

5.

Derived

Quantities.

All kinds

of

quantity

besides

the

fundamental

ones

just

mentioned

are

compounds

of the latter,

formed

by

multiplication

or

division,

such

as velocity,

accele-

ration,

momentum, work,

energy, moment,

power,

and

force-

distribution.

Some

of

these

are

merely

names

given for

convenience

to

certain combinations

of factors which

come

together

not

in

dealing with

first

principles,

but as

a result

of

common

algebraic

transformations.

6.

Homogeneous

Equations

are those

of such

a

form

that

they

are true for

any

arbitrary system

of

units, and

in

which

all

terms

combined

by

algebraic

addition

are

of

the same

kind.

Thus,

the

equation

8

-$

(in

which

g

=

the

acceleration

of

gravity and

t

the

time

of vertical

fall

of a

body

in

vacuo,

from

rest) will give

the

distance

fallen through,

s,

whatever

units

be

adopted

for

measuring

time

and

distance.

But if

for


17/216

PRELIMINARY CHAPTER.

3

g

we

write

the

numerical

value

32.2,

which

it

assumes

when

time is

measured in

seconds

and distance in feet,

the

equation

s

=

16.

If

is

true for

those units alone, and

the

equation is not

of

homogeneous form.

Algebraic combination of homogeneous

equations should

always

produce

homogeneous

equations

;

if

not,

some

error has

been

made

in

the algebraic work. If any

equation derived or proposed

for practical use

is

not homogene-

ous,

an

explicit

statement

should

be

made in the context as

to

the

proper

units

to

be

employed.

7.

Heaviness.

By

heaviness

of a substance

is

meant

the

weight of a cubic

unit

of

the

substance.

E.g.

the heaviness

of

fresh

water is

62.5,

in case the unit of force

is

the

pound,

and

the foot

the

unit

of

space; i.e.,

a

cubic

foot of fresh

water

weighs

62.5 lbs.

In

case

the

substance

is

not

uniform

in

composition, the

heaviness

varies

from

point

to

point.

If

the

weight

of

a

homogeneous

body

be denoted

by

G,

its

volume

by

V,

and

the

heaviness of

its

substance

by

y,

then

G

=

Vy.

Weight

in

Pounds

of a Cubic

Foot

(i.e.,

the

heaviness)

op various

Materials.

Anthracite,

solid

100

broken

57

Brick,

common

hard

125

soft

100

Brick-work,

common

112

Concrete

125

Earth,

loose

72

as

mud

102

Granite

164

to

172

Ice

58

Iron,

cast

450

wrought

480

Masonry,

dry

rubble

138

dressed granite

or

limestone 165

Mortar

100

Petroleum

55

Snow 7

wet

15 to

50

Steel 490

Timber

25 to

60

Water,

fresh 62.5

sea

64.0

8.

Specific

Gravity

is

the

ratio

of

the

heaviness

of

a

material

to

that

of

water,

and

is

therefore

an

abstract number.

9.

A

Material

Point

is

a

solid body,

or small particle,

whose

dimensions

are

practically

nothing,

compared

with

its range of

motion.


18/216

4 MECHANICS

OF

ENGINEERING.

10.

A

Rigid

Body is a

solid,

whose

distortion or

change

of

form

under any system of

forces

to

be

brought

upon

it

in

practice is,

for

certain purposes,

insensible.

11.

Equilibrium.

When

a

system of

forces

applied to a

body

produces

the

same effect as

if no force acted, so

far as

the

state

of

motion of

the

body

is

concerned, they are

said

to

be

balanced,

or to

be

in

equilibrium.

12.

Division

of the Subject.

Statics will treat

of

bodies at

rest,

i.e., of

balanced forces or

equilibrium;

dynamics,

of

bodies in

motion

;

strength

of

matenals

will

treat

of the

effect

of

forces in

distorting

bodies

; hydraulics, of

the

mechanics

of

liquids

;

pneumatics,

of the

mechanics

of

gases.

13.

Parallelogram

of Forces.

Duchayla's Proof.

To

fully

determine

a

force

we

must have given

its

amount,

its

direc-

tion,

and

its

point of application

in the

body.

It is

generally

denoted

in

diagrams

by

an arrow.

It is a matter

of

experience

that

besides

the

point

of

application

already spoken

of any

other may

be

chosen

in the

line

of action of

the

force.

This

is

called

the

transmissibility

of force

;

i.e.,

so

far

as

the

state

of

motion

of the

body

is

concerned,

a

force

may

be

applied

any-

where

in its line of action.

The

Resultant

of

two

forces (called

its

components)

applied

at a.

point

of

a

body

is a single

force

applied at the same

point,

which

will

replace

them. To prove

that this resultant

is given

in

amount

and position by the diagonal

of

the

parallelogram

formed

on

the two given

forces

(conceived

as. laid

off to some

scale,

so

many pounds

to

the inch,

say),

Duchayla's

method

requires

four

postulates, viz.

:

(1)

the resultant

of two forces

must

lie

in

the

same

plane

with

them

;

(2)

the

resultant

of two

equal forces

must

bisect the

angle

between them

;

(3)

if one

of

the

two

forces

be

increased,

the

angle

between

the

other

force

and

the

resultant

will

be

greater

than

before

; and

(4)

the

trans

missibility

of

force,

already

mentioned.

Granting

these,

we

proceed

as

follows (Fig.

1)

:

Given

the

two

forces

P

and

Q

=


19/216

PRELIMINARY CHAPTER.

5

P'

+

P

(P'

and

P

being

each

equal

to

P,

so

that

Q

=

2P),

applied

at 0.

Transmit

P

to

A.

Draw

the

parallelograms

OP and

AD; OP

will

also

be

a

parallelogram.

By

postulate

(2),

since

OB

is

a

rhombus,

P

and

P'

at

may

be

replaced

by

a

single

force

R'

acting through

B.

Transmit

R'

to B

and

replace

it

by

P

and

P'.

Transmit

P

from

B

to

J.,

P'

from

B

to

P.

Similarly

P

and

P ,

at

^4,

may be

replaced

by

a

single

force

R

passing

through

P;

transmit it there and re-

solve it into

P

and

P .

P'

is already

at

P.

Hence

P

and

P'

+

P ,

acting at

P,

are

equivalent

to

P

and

P'

+

P

act-

ing

at

0,

in

their

respective directions.

Therefore the

result-

ant

of

P

and

P'

-j-

P

must lie

in

the

line

OP,

the

diagonal

of

the

parallelogram

formed

on

P

and

Q

=

2P

at 0.

Similarly

C/

F?

/B

Fig.

1.

this

may be

proved

(that the

diagonal

gives

the

direction of

the

resultant)

for

any two

forces

P

and raP;

and

for

any two

forces

nP

and mP,

m

and n

being

any

two

whole numbers,

i.e.,

for

any two

commensurable

forces.

When

the forces are

incommensurable

(Fig.

2),

P

and

Q

being

the

given

forces,

we

may use

a

reduetio

ad

absurdum,

thus

:

Form

the

parallelo-

gram

OP

on

P

and

Q

applied

at

0.

Suppose

for

an

instant

that

R

the

resultant

of

P

and

Q

does not

follow the

diagonal

OP,

but

some

other

direction,

as

OP'.

Note

the

intersection

H,

and

draw

HG

parallel

to PB.

Divide

P

into

equal parts,

each

less

than

HP

;

then

in

laying

off

parts

equal

to

these

from

along

OB,

a

point

of

division

will

come

at

some

point

F

between

C

and B.

Complete

the

parallelogram

OFEO.

The

force

Q =

OF

is

commensurable

with

P,

and hence

their


20/216

6

MECHANICS

OP

ENGINEERING.

resultant

acts

along

OE. Now

Q

is

greater than Q ,

while

P

makes

a

less angle

with

P

than

OE,

which

is

contrary to

pos-

tulate

(3);

therefore

R

cannot

lie

outside

of

the

line

OD.

Q.E.D.

It still

remains

to

prove

that the

resultant

is

represented

in

a/mount,

as well as position,

by

the

diagonal.

OD

(Fig.

3)

is

\/r'

the direction of

P

the resultant

of

P

and

/F\

Q;

required

its

amount.

If

P'

be

a force

/rv

7\

7

equal

and

opposite

to

P

it

will balance

P

\

/

\q/

an

^

Q

5

i-

e

>

tne

resultant

of

P'

and

P

p

(j'a

must lie

in the

line

QO

prolonged

(besides

FlG

'

3

'

being

equal to

Q).

We can therefore

de-

termine P'

by

drawing

PA

parallel to

DO

to intersect

QO

prolonged

in

A

;

and

then

complete

the

parallelogram on

PA

and

PO.

Since

OFAP

is a

parallelogram P' must

=PA,

and

since

OAPD

is

a

parallelogram

BA=OD;

therefore

K'= OD

and also

P= OD.

Q.

E. D.

Corollary.

The resultant

of three

forces

applied

at

the

same

point

is

the

diagonal

of

the parallelopiped

formed

on the

three

forces.

14.

Concurrent

forces are those

whose

lines

of

action

intersect

in a

common

point,

while

non-concurrent

forces

are

those

which

do

not

so

intersect

;

results

obtained

for

a system

of

concurrent

forces are

really

derivable,

as particular

cases,

from

those per-

taining to a

system

of

non-concurrent

forces.

15.

Resultant.

A

single force, the

action

of

which,

as

re-

gards

the

state

of

motion

of

the

body acted

on, is

equivalent

to

that

of a

number of

forces forming

a

system,

is

said

to

be

the

Resultant

of that system,

and may

replace

the

system

;

and

con-

versely

a force which

is equal

and

opposite

to

the

resultant

of

a

system

will

balance

that system, or,

in

other

words,

when

it

is

combined

with that

system

there

will

result

a

new

system

in

equilibrium.

In

general,

as will

be seen,

a

given

system

of

forces

can

al-


21/216

PRELIMINARY

CHAPTER.

7

ways

De replaced

by

two

single

forces,

but

these

two

can

be

combined into

a

single resultant

only

in

particular

cases.

15a. Equivalent

Systems

are

those

which may be

replaced

by

the same

set

of

two

single forces

or,

in other

words,

those

which have

the

same

effect,

as

to state of

motion,

upon

the

given

body.

15b.

Formulae.

If in Pig.

3

the

forces

P

and

Q

and the angle

a

=

PO

Q

are

given,

we

have,

for

the

resultant,

BOB

=

^

P*

+

Q'

+

2 PQ

cos

a.

(If a is

>

90

its

cosine

is negative.)

In

general, given

any three parts

of either

plane

triangle

OBQ,

or OB B,

the other three may be obtained

by

ordinary trigonometry. Evidently if a

=

0,

R

=

P

+

Q;

if

a

=

180,

B

=

P-

Q;

and

if a

=

90,

B-

V

P

a

+

Q\

15c.

Varieties Of

Forces.

Great

care

should

be used in

deciding

what

may

properly

be

called

forces.

The

latter

may

be

divided

into

ac-

tions by

contact, and

actions

at

a

distance.

If

pressure exists

between

two

bodies and

they are

perfectly

smooth

at

the

surface

of

contact,

the

pressure

(or

thrust, or compressive action),

of one

against

the other constitutes

a force,

whose

direction is

normal to

the

tangent

plane

at

any point

of contact

(a

matter

of

experience)

;

while

if

those

surfaces are

not

smooth

there

may also

exist

mutual

tangential actions or

friction.

(If

the

bodies

really form

a

continuous substance

at

the

surface

considered, these

tangential actions are

called

shearing

forces.)

Again,

when

a rod or

wire

is

subjected

to

tension,

any portion

of

it

is

said to exert

a pull or tensile

force

upon

the remainder

;

the ability to do this

depends

on the

property

of

cohesion.

The

foregoing

are examples

of actions by

contact.

Actions

at a distance

are exemplified

in the mysterious

attractions,

or

re-

pulsions,

observable

in

the

phenomena

of gravitation,

electricity, and

mag-

netism,

where the

bodies

concerned

are not necessarily in contact.

By

the

term weight we shall always

mean the

force

of

the earth's

attraction

on

the

body

in

question,

and

not

the amount

of matter in it.

[Note.

In

some common

phrases, such

as The

tremendous

force

of

a

heavy

hody

in

rapid motion,

the word

force

is

not

used

in a technical sense, but signifies

energy

(as ex-

plained in

Chap.

VI.).

The

mere

fact

that

a

hody

is

in

motion,

whatever

its

mass and

velocity,

does not

imply that

it is under

the action

of any force,

necessarily.

For instance,

at

any point in the path of

a

cannon

ball through the air,

the

only

forces

acting on it are

the

resistance of

the air

and

the attraction

of

the earth,

the latter having

a

vertical and

downward

direction.]


22/216

PART

I.-STATICS.

CHAPTER I.

STATICS OF A

MATERIAL

POINT.

16.

Composition

of

Concurrent

Forces.

A

system

of

forces

acting

on

a

material

point is

necessarily

composed

of

concurrent

forces.

Case

I.

All

the

forces in

One

Plane.

Let

be

the

material

point,

the

common point

of

application

of

all

the

forces

;

P

t

,

JP

a

,

etc., the

given

forces,

making

angles

oc

v

etc.,

with

the

axis

X. By

the

parallelogram

of

forces

P

l

may

be resolved

into

and

replaced

by

its

components,

.P,cos

a

acting

along

X,

and

P

r

sin

a

along

Y.

fig.

4.

Similarly

all

the remaining

forces

may

be

re-

placed

by

their X

and Y

components. We have

now

a new

system,

the

equivalent of

that

first given,

consisting

of a

set of

X

forces, having the

same line of application

(axis

X),

and

a

set of

Y

forces,

all

acting

in

the line

Y. The

resultant

of

the

X

forces being

their

algebraic

sum (denoted

by

2X)

(since

they

have the

same line of

application)

we have

2X

P

1

cos

a

l

-J-

P

t

cos

ot

t

-j-

etc.

=

2(P

cos

a),

and

similarly

2Y=

P,

sin

a

s

-\-

P

3

sin

a

t

-\-

etc.

=

2(P

sin

a).

These

two forces,

2X

and

27,

may be

combined

by the

parallelogram

of

forces,

giving

R

{/(2XY

-\-

(2Y)'

as the

single

resultant of the whole

system, and

its

direction

is

deter-

mined

by

the

angle

or;

thus,

tan

a

=

-^y-;

see Fig.

5. For

equilibrium

to

.exist,

P

must

=

0,

which

requires,

separately,


23/216

STATICS

OF

A MATERIAL POINT.

9

2X

=

0,

and

27-

(for

the

two

squares

(2X)'

and

(2

7 )

a

can

neither

of them

be

negative quantities).

Case

II.

The forces

having

any directions

in space,

but all

applied

at

0,

the

material

point.

Let

P

J}

P

2

,

etc., be the given

forces,

j

3

,

making

the angles

or,,

/?

and

y

respectively, with

three arbitrary axes,

X, Y,

and

Z

(Fig.

6),

at

right

angles

to

each other

and intersecting

at

0,

the

origin.

Similarly

let

a

v

fl

v

y

t1

be

the angles made

by _P

a

with these

axes,

and

so

on

for

all

the

forces.

By

the

parallelopiped

of

forces,

Pi

may

be replaced

by its

components.

Xi

=

Pi

cos a

u

Yi

Pi

cos

/?

and Z

x

=

P, cos

y

x

;

and

Fig.

6.

Fig.

7.

similarly

for all

the

forces,

so that

the

entire

system is

now

replaced

by

the

three

forces,

2X

=

Pi

cos

,

+

P

2

cos

a

t

-f

etc

2Y=Pi

cos

A

+

P,

cos

J3,

+

etc;

2Z

P

x

cos

Xj

+

P

a

cos

x,

+

etc

and finally by

the

single

resultant

B

=

V(2Xy

+

(27)'

+

(2Z)

T

,

Therefore,

for

equilibrium

we

must

have

separately,

2J=

0,

27=

0,

and

2Z

= 0.

P's

position

may

be

determined

by

its

direction

cosines, viz.,

2X

a

cos

a

=

-Q-

;

cos

/>

27

2Z

;cmy

=

17.

Conditions

of

Equilibrium.

Evidently, in dealing

with

a

system

of

concurrent

forces, it would

be

a

simple matter to


24/216

10

MECHANICS

OF

ENGINEERING.

replace any two

of the

forces by

their resultant

(diagonal

formed

on

them), then

to

combine this

resultant with

a

third

force,

and so

on

until

all the

forces had

been

combined, the

last

resultant

being

the

resultant

of

the whole

system.

The

foregoing treatment, however, is useful in

showing

that

for

equilibrium

of

concurrent

forces

in

a

plane only

two conditions

are necessary,

viz.,

2X

=

and

2Y

'

=

;

while in

space

there are

three,

2X

=

0,

2J =

0,

and

2Z

=

0.

In

Case I.,

then,

we

have

conditions enough

for

determining

two

unknown

quantities

;

in

Case

II., three.

18.

Problems

involving equilibrium

of

concurrent

forces.

(A

rigid

body

in

equilibrium under

no

more

than

three

forces

may

be

treated

as

a material

point,

since

the

(two

or) three,

forces

are

necessarily

concurrent.)

Peoblem

1.

A

body weighing

G

lbs.

rests on

a horizontal

table

:

required the

pressure

between

it and the

table. Fig.

8.

Consider

the

body

free,

i.e.,

conceive

all other

bodies

removed

,

(the table

in this

instance),

being

replaced

by

the

1

forces which

they

exert

on

the

first

body.

Taking

G

the axis

^vertical

and

positive

upward,

and

not

0,

-+X assuming in

advance

either

the

amount

or

direc-

Tjq

tion

of

JV, the

pressure

of

the

table

against

the

body,

but

knowing

that

G,

the

action

of

the earth

on the

body, is

vertical

and

downward,

we have

here a

system of concurrent forces

in

equilibrium,

in

which

the

X

and

Y

components

of

G

are

known

(being

and

G

respectively), while

those,

iT

x

and

iT

T

,

of

N

are

unknown.

Putting

2X

0,

we

have

_ZT

X

+

=

;

i.e.,

iVhas

no

hori-

zontal

component,

..

_ZV

is

vertical.

Putting

2Y=0,

we

have

2F

T

G

=

0,

.-.

JV

T

=

-\-

G

;

or the

vertical

component

of

N,

i.e.,

N

itself,

is

positive

(upward

in

this

case),

and

is

numerically

equal

to

G.

Peoblem

2.

Fig.

9.

A

body

of weight

G

(lbs.)

is

moving

in a

straight

line over

a rough

horizontal

table

with

a

uniform

velocity

o

(feet

per

second)

to

the right.

The

tension

in

an

oblique

cord

by

which

it

is

pulled

is

given,

and

=

P

(lbs.),


25/216

STATIC8

OF A

MATERIAL

POINT.

11

which

remains

constant,

the

cord

making

a

given

angle

of

elevation,

a,

with

the

path

of

the

body.

Required

the

vertical

pressure

JV

(lbs.)

of the

table,

and

also

its

+

y

horizontal

action

F

(friction)

(lbs.)

against

the

body.

Referring

by

anticipation

to Newton's

first

law

of motion,

viz.,

a material

point

acted

on

by no force

or

by

balanced

forces

is

either

Fl

- 9 -

at

rest

or

moving

uniformly

in a straight

line,

we

see

that this

problem is

a

case

of balanced

forces, i.e.,

of

equilibrium.

Since

there

are

only

two unknown

quantities,

N and

F, we

may

determine them

by

the

two

equations

of

Case

I.,

taking

the

axes

X

and

Y

as

before.

Here let

us leave

the

direction

of

iVas well

as

its amount

to

be

determined

by

the

analysis.

As

F

must

evidently point

toward

the

left,

treat

it

as

negative

in

summing

the

X

components

;

the

analysis,

therefore,

can

be

expected

to

give only

its numerical

value.

2X

=

gives

P

cos

a

F= 0.

.-.

F

'

=

P

cos

a.

^JT^Ogivesir+Psina:-

G

=

0.

.-.

1ST

=

G

P

sin

a.

.'.

iV is

upward

or

downward

according

as

G

is

>

or

' '

W

while,

if

the

body

is

homogeneous,

y

is

the

same

for

all

its

ele-

ments,

and

being

therefore

placed

outside

the

sign

of

summa-

tion, is cancelled

out,

leaving

for

homogeneous

bodies

(

V

de-

noting the

total volume)

-

fxdV

-

fydV

-

fzdY

*

--^r ,

v

=

v->

and

s

--V--

(

2

)


33/216

PARALLEL

FORCES

AND

THE

CENTRE

OP

GRAVITY.

19

Corollary.

It

is

also

evident

that

if

a

homogeneous

body

is

for

convenience

considered

as

made

up

of

several

finite

parts,

whose

volumes

are

V

V

etc.,

and

whose

gravity

co-ordinates

are

as,,

y

s

1

;

x

y

z,

;

etc.,

we

may

write

-

_

v*+

r+...

x

-

7;

+

r,

+

.

.

.

.

w

If

the

body

is

heterogeneous,

put

6*,

(weights),

etc.,

instead

of

V

etc.,

in

equation

(3).

If the

body

is

an

infinitely thin

homogeneous shell

of uni-

form

thickness

=

h,

then

dV= hdF(dF

denoting

an element,

and

F

the

whole area

of one surface)

and

equations

(2)

become,

after cancellation,

v-Ml.

T

-

fydF

-

i-Ml

u)

Similarly,

for

a

homogeneous wire

of constant

small cross-

section

(i.e..

a

geometrical line,

having weight),

its

length

being

s,

and

an

element of length ds,

we

obtain

m=

Jj^.-^.-^

^

(5)

It

is

often

convenient

to

find

the

centre

of

gravity

of

a

thin

plate

by

experiment,

balancing

it on

a

needle-point;

other

shapes

are

not so

easily

dealt

with.

24.

Symmetry.

Considerations

of

symmetry

of form often

determine

the

centre

of

gravity

of

homogeneous solids

without

analysis,

or

limit

it to

a

certain

line

or

plane.

Thus the centre

of

gravity

of

a

sphere, or

any

regular

polyedron,

is at

its

centre

of

figure

;

of

a right

cylinder,

in

the

middle

of

its

axis

;

of

a

thin

plate

of

the

form

of a

circle

or

regular

polygon, in

the

centre

of

figure

;

of a

straight

wire

of

uniform

cross-section, in

the

middle of

its length.

Again,

if

a

homogeneous

body is

symmetrical about a

plane,

the

centre

of

gravity

must

lie in

that

plane,

called a

plane

of


34/216

20

MECHANICS

OP

ENGINEERING.

gravity

;

if

about a

line,

in

that

line

called

a

line

of

gravity

if about

a point, in that point.

25. By

considering

certain

modes

of subdivision of a

homo-

geneous body, lines or

planes

of

gravity are often

made appar-

ent.

E.g.,

a line

joining

the

middle

of

the

bases

of a

trape-

zoidal plate is a

line

of

gravity, since

it bisects

all

the

strips

of

uniform

width

determined

by

drawing parallels

to

the

bases

;

similarly,

a

line

joining

the

apex

of

a

triangular

plate to

the middle

of

the

opposite side is a

line

of

gravity. Other

cases

can

easily

be

suggested

by

the

student.

26.

Problems.

(1)

Required

the position of

the

centre

of

A

gravity

of

a,

fine

homogeneous wire

of

the

-j-3

form

of

a

circular arc,AB, Fig.

16.

Take

\

the origin

at the centre of

the

circle,

and

the

axis

21

bisecting

the

wire.

Let

the

u

\

A

dx j

length

of

the

wire,

s,

=

2s, ;

ds

=

ele-

^v

-

?//

1

mentofarc.

We need determine

only the

%/

x,

since

evidently

y

=

0.

Equations

(5),

,

fsads

fig.

16.

g

23,

are

applicable

here, i.e., x

.

s

From similar

triangles

we

have

ds

: dy

::

r

: x; .:

ds

rdy

x

,y

=

+

r n

y

**

zra

:.x

=

/

dy

q

,

i.e.,

=

chord

X

radius

length

of

wire.

For

a semicircular wire, this reduces to

x

=

2r

-5-

it.

Peoblem

2.

Centre

of

gravity

of

trapezoidal

{and

trian-

gular)

thin

plates,

homogeneous,

etc.

Prolong

the

non-parallel

sides

of

the

trapezoid

to

intersect

at

0,

which

take

as an origin,

making the

axis

X

perpendicular

to

the

bases

and

b

t

.

We

may

here

use equations

(4),

23,

and may take

a

vertical

strip

for

our

element

of

area,

dF,

in

determining

x

;

for

each

point

of

such

a

strip

has the

same

x.

Now

dF

=

(y

-\-

y')dx, and


35/216

PARALLEL

FORCES

AND

THE

CENTRE OF GRAVITY.

21

from similar triangles

y

-f-

y'

=

j

x.

Hence

F

= w

{bh

bji^)

can be

written -^

7

(h*

2A

h'),

and

x = -p

becomes

b

ph

hJiH

x'dx

1 I

+

-

7

W-

K)

=

-

J

'

2

A

2

U

3

A

a

~

LA

for

the

trapezoid.

2

For a

triangle A,

=

0,

and

we bave x

=

h; that

is,

the

o

centre

of

gravity

of

a

triangle

is

one third the

altitude

from

the

base.

The

centre of gravity

is

finally determined

by

knowing

Fig.

17.

Fio.

18.

that

a

line

joining

the

middles

of

I

and

b

l

is a

line of

gravity

or

joining

and

the

middle

of

b

in the

case

of a

triangle.

Problem

3.

Sector

of

a

circle.

Thin plate,

etc.

Let the

notation,

axes,

etc.,

be

as

in

Fig.

18.

Angle

of

sector

=

2a;

x

=

?

Using

polar

co-ordinates,

the

element of

area

dF

(a

small

rectangle)

=

pd


36/216

22

MECHANICS

OP

ENGINEERING.

(Note

on

double

integration.

The

quantity

cos

q>

J

p'

dp

\dq>,

is

that

portion

of the

summation / /

cos

q>p

,

dpd(p

which

belongs

to

a single

elementary

sector

(triangle),

since

all its

elements

(rectangles),

from

centre

to

circumference,

have

the

same

.)

That

is,

-

1

/*

P+

a

r* r+

2

r

sin

a

a

4

r

sin

fl

or,

putting

p

= 2a =

total angle of

sector, a;

=

^

g

4t*

For a

semicircular plate

this' reduces to

x

=

^

.

[iTofe.

In

numerical

substitution the

arcs

and

yS

used

above

(unless

sin or

cos

is

prefixed)

are understood

to

be

ex-

pressed

in

circular

measure (^-measure)

; e.g.,

for

a

quad-

rant,

/?

=

|

=

1.5707

+

;

for

30,/?

=

^;

or,

in

general, if

/J

.

,

180

,

n~\

In

degrees

=

,

then

p

in

^-measure

^\\

Problem

4.

Sector

of

a

flat

ring

;

thin

F A*

P^te)

e*

c

-

Treatment similar

to

that

of

^


37/216

Parallel

forces

and

the

centre

of

gravity.

23

Peoblem

5.

Segment

of

a

circle;

thin

plate,

etc.

Fig.

20.

Since

each

rectangular

element

of

any

ver-

Y

tical

strip

has

the

same

x,

we may

take

the

strip

as

dF

in

finding

x,

and

use

y

as the

half-height

of

the

strip.

dF

=

2ydx, and

from

similar

triangles

x

:

y

:

:

( dy)

:

dx,

i.e.,

xdx =

ydy.

Hence

from

eq.

23,

fatydx

_

2fiy'd

y

F

~

F

(4),

ft-

but

a

= the

half-chord,

hence,

finally,

x

=

^nr^-.

Problem

6.

Trapezoid; thin

plate,

etc.,

by

the

method

in

the corollary

of

23

; equa-

tions

(3).

Required

the

distance

x

from

the

base

AB.

Join

DB, thus

dividing

the

trape-

zoid

ABCD into

two

triangles

ADB

=

F

l

and

DBC

=

F

whose

gravity

a;'s are, re-

spectively,

x

x

=

ih

and

x,

=

|A.

Also,

F

l

=

ihb^

F,

=

%hb

and

F

(area of

trape-

zoid) =

A(&,

+

b,). Eq.

(3)

of

23

gives

Fxj=

F

1

x

1

-f-

F

t

x,

;

hence,

substituting,^,

-f-

\)x=

1

h+%bji.

Fio. 81.

#=

o

A

(i,

+

2ft.)

k

+ K

The

line joining

the

middles of

5,

and

5

is a line

of gravity,

and

is

divided in

such

a

ratio

by

the

centre

of

gravity

that

the fol-

lowing construction

for

finding

the

latter

holds

good

:

Prolong

each

base,

in opposite

directions,

an

amount equal

to

the

other

base;

join

the two

points thus

found:

the

intersection with

the

other line of

gravity is

the

centre

of

gravity

of

the

trape-

zoid.

Thus,

Fig.

21,

with BE= b,

and

DF=

l

join

FF,

etc.


38/216

24

MECHANICS

OF

ENGINEERING.

Problem

7. Homogeneous

oblique

cone

or

pyramid

Take

the

origin

at

the

vertex,

and

the

axis X

perpendicular

to

the

base

(or

bases,

if a

frustum).

In

finding

x

we

may

put

dY

volume

of

any

lamina

parallel

to YZ,

F

being

the base

of

such

a

lamina,

each

point of

the

lamina

having

the

same x.

Hence,

(equations

(2),

23),

but

and

x= ^/xdY,

Y=fdY=fFdx;

F:F,::x*:h,\

.:F=^o?,

For a frustum,

x

=

~r.-r-

s

t~

;

while

for a

pyramid, A,,

be-

4

/fc

2

/^

3

ing

=

0,

x

-rh.

Hence

the centre

of

gravity

of

a

pyramid

is

one

fourth

the

altitude

from

the base.

It

also lies in

the

line

joining

the

vertex

to the

centre

of

gravity

of

the base.

Problem

8.

If

the heaviness

of

the ma-

.,

-,

,

terial

of

the

above

cone

or pyramid varied

_//

directly

as

x,

y,

being

its

heaviness at

the

fig. 22.

base

F

we

would

use equations

(1),

23,

putting

y

=

-r

x

;

and finally,

for

the frustum,

ft,

-

4

h:-hs

x=

=-.

5 h:

-

k

*

)

4

and

for

a

complete

cone

x

=

h

t

.

27.

The

Centrobaric

Method.

If

an

elementary

area

dF

be

revolved

about

an

axis in

its

plane,

through

an

angle

a

and

are

there-

fore

(

34)

equivalent

to

a

single

couple,

in

the

same

plane

with

a

moment

=

(7&XJ/.).

Treating

all

the

remaining

forces

in

the

same

way,

the

whole

system

of

forces

is

replaced

by

the

force

2(X)

=X

X

+

X,

+

...

at

the

origin,

along

the

axis

X;

the

force

2(7)

-

7

V

+

7,

+

...

at

the

origin,

along the

axis

T;

and

the

couple

whose

mom.

G

=

2

(

Yx

Xy),

which

may be

called

the couple

(see

Fig.

32),

and

may be

placed

anywhere

in

the

plane.

Now

2(X)

and

2(

7)

may

be

combined

into

a

force

E ;

i.e.,

.

2X

E

=

Y(2Xy

-+-

2

7f

and its

direction-cosine

is

cos

a

=

rr-

Since,

then,

the

whole

system

reduces

to

C

and E,

we

must

have

for

equilibrium

E

0,

and

G

=

;

i.e.,

for

equilibrium

2X=

0,

27=

0,

and 2(7x-Xy)

=

0.

.

eq.

(1)

If

R

alone

=

0,

the

system

reduces to a couple

whose

mo-

ment is

G

=

2(

7xXy)

;

and if

G

alone

=

the

system

re-

duces to

a

single force

E,

applied at the origin. If, in

general,

neither E

nor

G

=

0,

the

system

is

still

equivalent

to

a

single

force, but

not

applied

at

the

origin (as could

hardly

be

ex-

pected,

since the

origin

is arbitrary)

;

as follows

(see Fig.

33)

:

Replace

the

couple

C

by one of equal moment,

G,

with

each

force

=

E.

Its arm will therefore

be

-^.

Move this couple

in the

plane

so that

one of its

forces

E

may

cancel the

E

al-

ready

at

the origin,

thus

leaving

a

single

resultant

E

for the

wljple system, applied

in a

line

at a

perpendicular

distance,

c

=

-55

,

from

the

origin,

and

making

an

angle

a

whose cosine

=

2X

.

,

,

.

v

~s~,

with the axis

X.

36.

More convenient

form

for

the

equations

of

equilibrium

of

non-concurrent

forces

in

a

plane.

In

(I.),

Fig.

34, O

being


47/216

STATICS

OF

A

BIGUD BODY.

33

any

point

and a

its

perpendicular

distance

from

a force

P

put in

at two

equal

and

opposite forces

P

and

P'

=

and

||

to

P,

and

we

have

P

replaced

by

an

equal

single

force

P'

at

O,

and

a

couple

whose moment

is

+

Pa. (II.)

shows

a

simi-

lar

construction,

dealing

with

the Jf

and ^components

of

P,

so

that in

(II.) P

is

replaced

by

single

forces

X'

and

Y'

at

/P v. -am

Fio.

33.

(and

they

are

equivalent

to

a

resultant

P',

at

0,

as

in

(I.), and

two

couples

whose

moments

are

-|-

Yx

and

Xy.

Hence,

being

the

same

point in both

cases,

the couple

Pa

is

equivalent

to

the

two

last

mentioned,

and,

their

axes

being

parallel, we

must

have

Pa

=

Yx

Xy.

Equations

(1),

35,

for

equilibrium,

may now

be

written

2X

0,

2

Y

=

0,

and

2{Pa)

=

0.

. .

(2)

In

problems involving

the

equilibrium of

non-concurrent

forces in

a

plane,

we have

three

independent

conditions,

or

equations,

and can determine at most

three unknown

quantities.

For

practical

solution,

then,

the

rigid

body

having been

made

free

(by

conceiving the actions

of

all

other

bodies

as

repre-

sented by

forces), and

being

in

equilibrium (which

it

must

be

if

at

rest),

we

apply equations

(2)

literally

;

i.e.,

assuming an

origin

and

two

axes,

equate

the

sum of the

JT

components

of

all

the

forces to zero

;

similarly for the

Y

components

;

and

then

for the

moment-equation,

having

dropped

a

perpen-

dicular

from

the origin

upon

each

force,

write the

algebraic

sum

of

the

products

{moments)

obtained

by

multiplying

each

force

by

its

perpendicular,

or

lever-arm, equal to zero,

call-

ing

each

product

+

or

according

as the ideal

rotation

ap-

pears'against,

or with,

the

hands of a watch,

as seen

from

the

same side

of the

plane. (The

converse

convention

would

do

as

well.)


48/216

34 MECHANICS

OP

ENGINEERING.

Sometimes

it is

convenient

to

use

three

moment

equations,

taking

a

new

origin

each time,

and

then the

2X=

and

2

Y

=

are

superfluous,

as they

would

not

be

independent

equa-

tions.

37.

Problems

involving

Non-concurrent

Forces

in a

Plane.

Remarks.

The

weight

of a

rigid body is

a

vertical

force

through

its centre of

gravity,

downwards.

If the

surface

of

contact of two

bodies is

smooth the

action

(pressure, or force)

of one

on

the

other

is

perpendicular

to

the

surface at the

point

of contact.

If a

cord

must

be

imagined

cut, to

make

a

body

free, its

tension

must

be

inserted

in

the

line

of the cord,

and

in such

a

direction as to

keep

taut the

small

portion

still fastened to

the

body.

In

case

the' pin

of

a

hinge must

be

removed,

to

make

the

body free, its pressure

against

the

ring

being

unknown

in

direction

and

amount,

it is

most

convenient

to

represent it

by

its

unknown

components

X

and

Y,

in known

directions.

In

the

following

problems

there

is

supposed to

be

no

friction.

If the line

of

action

of

an

un-

known

force

is

known,

but

not

its direction

(forward or

back

ward),

assume

a

direction for

it

and

adhere

to it

in

all

the

three

equations,

and if

the

assumption is

correct

the

value of

the

force, after elimination, will

be

positive

;

if

incorrect,

negative.

Problem

1.

Fig.

35.

Given

an

oblique

rigid rod,

with

two

loads

G

t

(its

own weight)

and

G

t

;

required

the

reaction

of the

smooth

vertical

wall at

A,

and

the

direction

and

amount

of

the

SAin^-pressure

at

0.

The

reaction

at

A

i

'

a

9Y

...

//

'

must

be

horizontal

;

call it

X. The

pres-

jj-j

sure

at

0,

being

unknown in

direction,

will

have

both

its

X

and

Y components

un-

known.

The

three

unknowns,

then,

are

X

,

X,

and

Y

while

G G

a,,

a

and

h

are

known.

The

figure

shows

the

rod

as

a

free

body, all the

forces

acting

on

it

have been put in, and, since

the

rod is at

rest,

constitute

a

sys-

tem

of

non-concurrent

forces

in

a

plane,

ready

for

the

condi-

tions

of

equilibrium.

Taking origin

and

axes

as

in

the

figure,


49/216

STATICS

OF A RIGID

BODY.

35

2X

=

gives

+X

-

X'

=

;

2Y

=

G,

=

0;

while

2

(Pa)

=

0,

about

G

k

a

t

G^a

t

=

0.

(The

moments

of

each,

= zero.)

By eliminationre,

G,;X

e

=

X-,--[G

l

a

l

+G>a

1

]

gives

+ Y

a

-

G

x

0,

gives

-\-

Xh

X

and

r

o

about

6>

we

obtain

Y

=


50/216

36

MECHANICS

OF

ENGINEERING.

First, required

the

reactions

of

the

supports

V

1

and

V

t

;

these and

the

loads are

called

the

external

forces. 2(Pa)

about

gives

Yfia

-

P, .

\a

-

P,

while

2(Pa)

about

K

= gives

la

0;

and

-

V

x

.

3a

+

P

s

.

\a

+

PJjfl

+

P^a

=

0;

F

=

i[5P, +

3P,

+

P

3

]

;

F

=

K-P,

+

3P.

+

5PJ.

Secondly,

required the stress

(thrust

or

pull,

compression

or

tension)

in each of the

pieces

A, B,

and Ccut

by

the

imaginary

line

PE.

The

stresses

in the

pieces

are

called

internal forces.

These appear

in

a system of

forces acting on a free

body

only

when

a

portion

of

the

truss

or

frame is conceived

separated

from the remainder

in such a way

as to expose

an

internal

plane of one

or

more

pieces. Consider

as a free

body

the por-

tion

on

the left

of

DE

(that on

the right would serve

as

well,

but

the pulls or

thrusts

in

A,

B,

and

would

be

found

to

act in

directions

opposite

to those

they

have

on

the

other

portion

;

see

3).

Fig.

39.

The

arrows

(forces)

A,

B,

and

C

are

not

pointed yet.

They, with

V

x

,

P

and

P

s

,

form

a system in

equilibrium.

H~

Fig. 39.

2(Pa)

about

O

=

gives

(Ah)

-

V$a

+

P, .

fa

+

P,

.\a

=

0.

Therefore

the moment

(Ah)

=

Ja[4

V

x

3P,

PJ,

which

is

positive,

since

(from

above) 4

7,

is

>

3P,

+

P

v

Hence

A

must

point

to

the

left,

i.e.,

is

a

thrust

or

compression,

and

is

^F.-SP-PJ.

Similarly,

taking

moments

about

0

the

intersection

of

A

and

B, we have an

equation

in which

the

only

unknown

is

C,

viz.,

(Oh)

-

Yfr

+

P

x

a

=

0.

.-.

(Oh)

=

a[3

F,

-

2P,],


51/216

STATICS

OF

A

RIGID

BODY.

37

a

positive

moment,

since

3

V

x

is

>2P,

;

.-.

Cmust

point

to

the

right, i.e., is

a

tension,

and

=

^3

V,

2P

Finally,

to

obtain

B,

put

2(vert.

comps.) =

0;

i.e.

(Pcos


52/216

38

MECHANICS

OE

ENGINEERING.

Considering

the

first

force

JP

replace

it

by

its

three com-

ponents parallel

to

the axes, X

l

=

I

3

1

cos

a^;

Y

1

=

JP

l

cos

yff,;

and

Z^

=

P

1

cos

y

l

(P

x

itself

is

not

shown in

the

figure). At

0,

and

also at

A,

put

a pair of

equal

and

opposite

forces,

each equal

and

parallel

to

Z,;

-Z, is

now replaced by

a

single

force

Z

x

acting upward at

the origin,

and

two

couples,

one

in

a

plane

parallel

to

YZ

and having

a

moment =

-Z,y,

(as

we

see

it

looking

toward

from

a

remote

point

on

the

axis

-\-

X),

the other in a

plane

parallel

to

XZ

and

having

a mo-

ment

= -|-

Z

t

x

t

(seen

from

a

remote

point

on

the

axis

-f-

Y).

Similarly at

and

O

put

in pairs

of

forces

equal and parallel

to X

and we have

X

at

B,

replaced

by the

single force

X

x

at the

origin,

and the couples, one

in

a

plane parallel to

XY

t

and

having

a

moment

-j-

X

t

y

seen

from

a

remote

point on

the

axis

-\-

Z,

the

other in a plane parallel

to

XZ,

and

of

a

moment

=X

1

s

1

,

seen

from

a

remote

point

on

the

axis

-\-Y\

and finally,

by

a

similar device,

Y

1

at

B

is replaced

by a

force

Y

1

at the origin and

two

couples,

parallel

to

the

planes

XY

and

YZ,

and having

moments

Y

t

x

1

and

+

Zjz,,

respective-

ly.

(In

Fig.

42

the single

forces at

the

origin are broken

lines,

while

the

two

forces constituting any one

of the six

couples

may

be

recognized

as

being

equal and parallel, of

opposite di-

rections,

and

both

continuous,

or

both

dotted.)

We

have,

therefore,

replaced

the

force

_P

a

by

three

forces X

v

,

Y

t

,

Z

at

0,

and

six

couples

(shown more clearly

in

Fig.

43;

the couples have

been

transferred to

symmetrical

posi-

tions). Combining

each

two

couples

whose

axes

are

parallel

to

X,

Y,

io. 43.

or Z,

they can

be

reduced

to

three, viz.,

one with

an

X

axis and

a

moment

=

T ,2,

Z

l

y

l

;

one with a

Zaxis

and

a moment

==

Z

x

x

x

X

x

z^\

one

with

a Z

axis

and

a

moment =

X,y

1

Y

x

x

v


53/216

STATICS OF A BIGID

BODY.

39

Dealing

with

each

of

the

other

forces

P

J

etc.,

in

the

same

manner,

the

whole

system

may finally

be

replaced

by

three

forces

2X,

2

7,

and 2Z,

at the

origin

and three couples

whose moments are,

respectively,

L

=

2(

Yz

Zy)

with its

axis parallel to

X

3f

=

2{Zx

Xz)

with its

axis

parallel

to

Y;

N

=

2(Xy

Yx)

with

its axis

parallel

to Z.

The

axes

of

these

couples,

being

parallel to

the

respective

co-ordinate

axes

X,

Y,

and

Z,

and

proportional

to

the

mo-

ments L,

M,

and N,

respectively,

the

axis

of

their

resultant

C,

whose

moment

is

G,

must

be

the

diagonal of

a

parallelo-

pipedon

constructed

on

the

three

component

axes

(propor-

tional

to) L, M,

and

N.

Therefore,

G

= VIF+jF+lF,

while

the

resultant

of 2X,

2

Y,

and

2Z is

b

=

V{zxy

+

(2

Yy

+

(2zy

acting

at

the

origin.

If

a,

(3,

and

y

are

the

direction-angles

2j

^

Y

~SZ

of

R,

we

have cos

a

=

-^-,

cos

fi

=

~^-,

and

cos

y

=

-^

;

while

if

A,

M,

an(

i

v

are

those

of

the

axis

of the

couple

G, we

L

M

3

N

have

cos

A.

=

77,

cos

p.

=

-q,

and

cos

v

=

-g.

For

equilibrium

we

have

both

G

=

and

R

=

;

i.e.,

separately,

six

conditions,

viz.,

2X=0,2Y

=

0,

2Z=0

;

and

2=0,

3/=0, i7=0

.

(1)

Now,

noting

that

2X

=0,27=0,

and

2{Xy

Yx)=0

are

the

conditions

for

equilibrium

of

the

system of

non-concur-

rent

forces

which

would

be

formed

by

projecting

each

force

of

our

actual

system

upon

the

plane

XY,

and

similar

relations

for

the

planes

YZ

and

XZ,

we

may restate

equations

(1)

in

another

form,

more

serviceable

in practical

problems,

viz.

:

Hote.

Xf

a

system

of

non-concurrent

forces

in

space is

in

equilibrium,

the

plane

systems

formed

by projecting

the

given

system

upon

each

of

three

arbitrary

co-ordinate

planes

will

also

be

m

equilibrium.

But

we

can

obtain only

six

independent


54/216

40

MECHANICS

OF

ENGINEERING.

equations in

any case,

available for

six unknowns.

If

H

alone

=

0,

we

have

the

system

equivalent to

a

couple

O,

whose

moment

=

G

;

if

G

alone

=

0,

the system

has a

single re-

sultant

It

applied at the

origin.

In

general,

neither

It

nor

G

being

=

0,

we

cannot

further

combine

H

and C(as was done

with

non-concurrent forces

in a

plane) to produce

a single

re-

sultant unless

It

and Care

in the

same

plane; i.e.,

when the

angle

between

R

and the axis

of

G

is

=

90.

Call

that

angle

6.

If,

then,

cos

6

=

cos

a

cos

A.

-j-

cos

ft

cos

jn

-f-

cos

y

cos

v

is

= =

cos

90,

we

may combine

It

and

C

to

produce

a

single

resultant

for

the

whole system

;

acting

in

a plane

con-

taining It

and parallel to the plane of

C

in

a

direction

parallel

to It,

at a

perpendicular

distance

c

=

-5

from

the

origin

and

=

It

in

intensity. The

condition that a system of

forces in

space

have

a

single

resultant

is, therefore, substituting

the

previously

derived values of the cosines,

(2J) . L

-J-

(2

Y)

.

M

+

(2Z)

.

N

=

0.

This

includes the cases when It is zero and

when

the

system

reduces to

a

couple.

To

return to

the general

case, It and

C

not being

in

the

same

plane,

the

composition

of forces

in

space cannot

be

further

simplified. Still

we

can

give any

value

we please

to

P,

one of

the forces of the couple

C,

calculate

the

correspond-

G

ing arm a

=

-p,

then

transfer

C

until one

of

the

_P's has

the

same

point of

application

as

It,

and combine them

by

the

parallelogram of forces. We thus

have

the

whole

system

equivalent

to

two forces, viz., the second

P,

and the

resultant

of It

and

the

first

P-

These

two

forces

are not

in

Documents

Church_statics and Dynamics 1886